GRE Physics GR 1777, Problem 002 Solution

You should know

If the velocities of the particles are $\vec{u_1}$ and $\vec{u_2}$ before the interaction, and afterwards they are $\vec{v_1}$ and $\vec{v_2}$, then

$$ m_1\vec{u_1} + m_2\vec{u_2} = m_1\vec{v_1} + m_2\vec{v_2} $$
$$ W = \Delta E_k = \Delta\frac{1}{2}mv^2 $$

Solution

By the linear momentum conservation,

  • x-direction
\begin{equation} \begin{array}{rcl} m_{1}u_{1,x} + m_{2}u_{2,x} & = & (m_{1} + m_{2})v_{x} \\ (0.5) \cdot 2 + (1.0)\cdot 0 & = & (0.5 + 1.5) v_{x} \\ \end{array} \end{equation}$$v_{x} = \frac{2}{3}$$
  • y-direction
\begin{equation} \begin{array}{rcl} m_{1}u_{1,y} + m_{2}u_{2,y} & = & (m_{1} + m_{2})v_{y} \\ (0.5) \cdot 0 + (1.0) \cdot 1 & = & (0.5 + 1.5) v_{y} \\ \end{array} \end{equation}$$v_{y} = \frac{2}{3}$$

Thus,

$$v = \sqrt{v^2_x + v^2_y} = \frac{2\sqrt{2}}{3}$$

Using the work-energy principle,

\begin{equation} \begin{array}{rcl} W = \ \ \Delta E_k & = & \Delta \frac{1}{2}mv^2 \\ & = & \frac{1}{2}((0.5) \cdot 2^2 + 1 \cdot 1^2) - \frac{1}{2}\left((1.5) \cdot \left(\frac{2\sqrt{2}}{3}\right)^2\right) & = & \frac{5}{6} J \\ \end{array} \end{equation}

If we assume that all energy generated by collision is converted to heat, then the heat energy produced by the collision is $\frac{5}{6} J$

Answer

(E) $\frac{5}{6}J$

 

 

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